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Good afternoon, I am trying to understand the basics of some quantum mechanics theorems (e.g. Uncertainity principle). I'm looking for the correct way to read this expression while I'm speaking. For instance $\langle a|b\rangle$ or

$$\langle(A-\langle A\rangle) \psi|(A-\langle A\rangle)\psi\rangle$$

And, more in general, what is the basic difference between $\langle a|$ and$|a\rangle$?

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    $\begingroup$ I think this question is just too broad: the answer is a chapter in any QM book. I particularly recommend Shankar's. $\endgroup$ – Javier Sep 2 '19 at 13:55
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    $\begingroup$ Did you have a look at the Wikipedia page of bra-ket notation? Does this answer your question or can you re-formulate it to be more specific? $\endgroup$ – ahemmetter Sep 2 '19 at 13:56
  • $\begingroup$ Many thanks. I looked at it but I have some doubt on how to properly read with my voice the longest expression I posted. $\endgroup$ – muserock92 Sep 2 '19 at 13:57
  • $\begingroup$ Wait, do you mean how to read out loud the expressions? I'm not sure that's really a physics question; your last question about the difference between bras and kets is one, but it's probably too broad for the site. $\endgroup$ – Javier Sep 2 '19 at 14:12
  • $\begingroup$ I was thinking about how to tell this expression if I have to speak with someone. $\endgroup$ – muserock92 Sep 2 '19 at 14:15
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$\newcommand{\ket}[1]{\left|#1\right>}$ $\newcommand{\bra}[1]{\left<#1\right|}$ Not quite sure I understand the entire question, but you can think of quantum systems as vectors:

  • $\ket{A}$ is a column vector containing the probability amplitudes of the system
  • $\bra{A}$ is a row vector, and the complex transpose of $\ket{A}$, i.e. $\bra{A} = \ket{A}^{\dagger}$. It comes in handy when we want to calculate the expected value of a measurement outcome. See it as a mathematical tool.
  • $\left<A| B\right>$ is the inner product of $A$ and $B$, which gives you a scalar
  • $\ket{A}\bra{B}$ is the outer product of $A$ and $B$, which gives you a matrix (called the density matrix)

For two-level quantum systems such as a qubit, this way of thinking is very intuitive, since $\ket{A}$ simply becomes a column vector with 2 entries. However, when the systems are continuous, we suddenly deal with infinite dimensional vectors but essentially the logic is the same.

Hope this helps.

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  • $\begingroup$ While correct, muserock92 seems more concerned with the verbal expression, rather than the meaning of the terms. $\endgroup$ – Kyle Kanos Sep 4 '19 at 11:55

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